Projectile Motion and Angle: Solving for Optimal Launch Angle to Double Range

[terryaki]

I need help putting this problem into a workable equation:

A projectile is launched from ground level at an angle of 12 degrees above the horizontal. It returns to ground level. To what value should the launch angle be adjusted, w/o changing the launch speed, so that the range doubles?

So far I tried this:

I broke the problem into its components: x & y.
FOR X: {x=Vo*COS(12)t Vox=Vo*Cos(12) a=0 t=t}
FOR Y: {y=? Voy=Vo*SIN(12) a=-9.8 m/s^2 t=t}

Then I used t=x/Vo*COS(12), then substituted that for T in the Y parts, so that:

y= tan(12)x - [(4.9 x^2)/(Vo^2*cos(12)^2)]

but then I got: x=tan(12)y, which doesn't help me at all.

I'm guessing I'm approaching this problem in a totally WRONG way!

 

[dduardo]

so you start out with a vector v-> and an angle theta1.

x = |v->| * cos(theta1) * t
y = |v->| * sin(theta1) * t - (1/2) * g * t^2

when y = 0, the projectile will be on the ground

0 = |v->| * sine(theta1) * t - (1/2) * g * t^2

t = 0 , 2 *|v->| * sin(theta1) / g

when t = 0 is at the start, so the other solution is the time it hits the ground

x = |v->| * cos(theta1) * 2 *|v->| * sin(theta1) / g

pluging in t into the x equation and doing some math you get:

x = |v->|^2 * sin( 2 * theta1) / g

If you want the range to be double then it is 2 times the x equation with the new angle.

x2 = 2 * |v->|^2 * sin( 2 * theta2) / g

equate x and x2 to find theta2. You know theta1 = 12 degrees

x = x2 , then theta2 = 5.87 degrees

And there is your answer

 

[M98Ranger]

Your approach to breaking the problem into its components is correct. However, the equation you have derived for the y-component is not correct. The correct equation for the y-component of projectile motion is:

y = Voy*t + (1/2)*a*t^2

where Voy is the initial vertical velocity, a is the acceleration due to gravity (-9.8 m/s^2), and t is the time.

To solve for the optimal launch angle that will double the range, we need to consider the range equation:

R = Vox*t

where R is the range, Vox is the initial horizontal velocity, and t is the time.

We can substitute the equation for t that you have derived (t=x/Vo*cos(12)) into the range equation to get:

R = Vox * (x/Vo*cos(12))

We can then substitute the equation for Vox (Vox=Vo*cos(12)) to get:

R = (Vo*cos(12)) * (x/Vo*cos(12))

Simplifying, we get:

R = x

So, the range is directly proportional to the horizontal distance x. This means that if we want to double the range, we need to double the horizontal distance x.

Now, we can use the y-component equation to find the time t it takes for the projectile to reach the ground again. Setting y=0 and solving for t, we get:

0 = Voy*t + (1/2)*a*t^2

Solving for t, we get:

t = (2*Voy)/a

Substituting this value of t into the range equation, we get:

R = Vox * [(2*Voy)/a]

Since we want to double the range, we can set R=2R and solve for the optimal launch angle:

2R = Vox * [(2*Voy)/a]

Simplifying, we get:

2R = (Vo*cos(12)) * [(2*Vo*sin(12))/a]

Rearranging, we get:

2 = (cos(12)*sin(12))/a

Solving for a, we get:

a = (cos(12)*sin(12))/2

Now, we can use this value of a in the y-component equation to find the optimal launch angle. Setting y=0 and solving for the launch angle, we get:

0 = Voy*t + (1/2)*[(cos